The Secret Behind -1 Factorial

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Dive into the mystery of -1 factorial! Explore advanced math topics like integrals and complex numbers as we tackle this perplexing question. Meet the gamma function, the key to extending factorials beyond positive integers. But there's a twist—calculus shows the integral diverges to infinity. Is -1! factorial really infinity, or is it more complex? Unlock the secrets with analytic continuation and decode Wolfram Alpha's intriguing output.

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Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information. Viewers should always verify the information provided in this video by consulting other reliable sources.

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BriTheMathGuy

The infinity with a tilde is "complex infinity"; it's an infinity without a direction (the "north pole" on the Riemann sphere).
You get the same thing by typing 1/0 into WolframAlpha, since 1/0 is defined on the Riemann sphere.

ValkyRiver

I hoped you’d calculate a residue of the pole, or something. 🤔

degrees

How about a video about the third derivative of the gamma function evaluated at one, an how it relates to the appery's constant, the euler-mascheroni constant and Pi?

RGAstrofotografia

Of course the first pole of the analytic continuation of the gamma function occurs at e^iπ

TheLethalDomain

Sorry but I feel like it wasn’t very clear. What exactly does “complex infinity” mean from wolframalpha? Does it mean like the magnitude of the complex output grows unbounded as the distance between the input and -1 get closer? If someone could explain this I would greatly appreciate it.

Ninja

The main problem about this occurs where lets say you want to try (-n)!
But if its an even amount its result is positive and if its odd its result is also odd... thats one reason why (-n)! is undefined

huzefa

Another reason (and also simpler) why (-1)! is undefined:

We all know:
n! = n(n-1)(n-2)...(3)(2)(1)

But this can be expressed as:
n! = n(n-1)!

If we move (n-1)! to the left, we get:
(n-1)! = n!/n

For example:
n = 3
(3-1)! = 3!/3
2! = 6/3 = 2 ✅

n = 2
(2-1)! = 2!/2
1! = 2/2 = 1 ✅

n = 1
(1-1)! = 0!/1
0! = 1/1 = 1 ✅

If we want to find (-1)!, we substitute n = 0:
(0-1)! = 0!/0 ❗
(-1)! = 1/0 ❗

As you can see, getting (-1)! requires dividing by zero, which is undefined.

Zettabyte

You don't need the Gamma Function to go negative
n! = (n+1)!/(n+1) -> 0! = 1!/1 = 1 -> (-1)! = 0!/0
It's a bit of a mess but kind off the same

bjarnivalur

Not a fan of this one. Bri explained factorials and the gamma function a bit (cool) and then said (-1)! Is a special kind of infinity and we can talk a lot about it… then the video ends?

justrandomthings

At this point, you should change your name to BrilliantTheMathGuy

MatterOp

Other people have already given the derivation of (-1)! by the recurrence relation, so I'm going to ask a different question: Why does Γ(x) have t^(x-1) instead of just t^x? The minus 1 just feels so artificial and all it seems to do is push the gamma function _away_ from the factorial it's used to extend. There is an alternative function Π(x) which is defined for all complex numbers except negative integers, but also has Π(n) = n! for all natural numbers n, (so all numbers the traditional factorial is defined for) rather than (n-1)! for all positive numbers n. The factorials importance in calculus and combinatorics show no sign of a -1 and just use the factorial as is, so Π(x) would appear more natural when trying to extend them compared to Γ(x+1).
Is this question asked in many places? Yes. Have I ever seen a satisfying answer? No.

angeldude

After careful consideration I have decided to leave -1! Undefined for 2 reasons. First off we know (x-1)! Is x!/x. This is proof for 0! Being 1. But then for -1! We have 0!/0. 0! Is 1 so we have 1/0 and we don’t like that. Secondly, factorials can be considered the amount of possible arrangement of x items. You can arrange 2 items 2 ways (2!) 3 items 6 ways(3!) and 4 items 24 (4!). So how many arrangements can you arrange with -1 items? That doesn’t make a hint of sense. So i’ve Decided to leave it undefined.

Questiala

I tried this without the knowledge of gamma functions, and got pretty much? the same result. I used the fact that n!/(n-1)!=n, so 1!/0!=1, meaning 0!=1, and them I plugged in n=0. This gives the result that 0!/(-1)!=0, so (-1)!=1/0, which is undefined, but if you think factorial as a function and 1/0 as a limit, then it just blows up to infinity.

frog-dw